A 1.19-kg aluminum pot contains 2.31 L of water. Both pot and water are initially at \(19.7^{\circ} \mathrm{C} .\) How much heat must flow into the pot and the water to bring their temperature up to \(95.0^{\circ} \mathrm{C}\) ? Assume that the effect of water evaporation during the heating process can be neglected and that the temperature remains uniform throughout the pot and the water.

Short Answer

Expert verified
Answer: The total heat required is approximately 805,062.55 J.

Step by step solution

01

Finding the mass of water

Given the volume of water is 2.31 L, we need to convert it into mass. We know that the density of water is \(\rho = 1000 \frac{kg}{m^3}\) or \(1 \frac{kg}{L}\). Therefore, multiplying the volume by density, we can find the mass of water. $$m_{water} = \rho \times V = 1 \frac{kg}{L} \times 2.31 L = 2.31 kg$$
02

Determine the specific heat capacities

We need to find the specific heat capacities of aluminum and water. The specific heat capacity is a property of the material and given by: \(c_{Aluminum} = 903 \frac{J}{kg \cdot K}\) and \(c_{Water} = 4186 \frac{J}{kg \cdot K}\).
03

Calculate the change in temperature

We are given the initial temperature, \(T_{initial} = 19.7^{\circ} \mathrm{C}\), and the final temperature, \(T_{final} = 95.0^{\circ} \mathrm{C}\). Calculate the change in temperature for both the aluminum pot and water using the formula: \(\Delta T = T_{final} - T_{initial}\). $$\Delta T = 95.0^{\circ} \mathrm{C} - 19.7^{\circ} \mathrm{C} = 75.3^{\circ} \mathrm{C}$$
04

Calculate the heat required for the aluminum pot

Now we can use the equation for heat transfer to calculate the heat required to raise the temperature of the aluminum pot. $$Q_{Aluminum} = m_{Aluminum}c_{Aluminum}\Delta T = 1.19 kg \times 903 \frac{J}{kg \cdot K} \times 75.3 K$$ $$Q_{Aluminum} = 80594.57 J$$
05

Calculate the heat required for the water

Use the equation for heat transfer to calculate the heat required to raise the temperature of the water. $$Q_{Water} = m_{Water}c_{Water}\Delta T = 2.31 kg \times 4186 \frac{J}{kg \cdot K} \times 75.3 K$$ $$Q_{Water} = 724467.978 J$$
06

Calculate the total heat required

Finally, add the heat required for the aluminum pot and the water to get the total heat required. $$Q_{Total} = Q_{Aluminum} + Q_{Water} = 80594.57 J + 724467.978 J = 805062.548 J$$ The total heat required to bring the temperature of the aluminum pot and the water up to \(95.0^{\circ} \mathrm{C}\) is approximately \(805,062.55 J\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat Capacity
At the heart of understanding heat transfer in materials like the aluminum pot and water in our exercise is the concept of specific heat capacity. This property is crucial because it defines how much heat energy is needed to change the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin).

Imagine you're warming up different materials—each will require a different amount of energy to reach the same temperature based on their specific heat capacity. Metals like aluminum usually have a lower specific heat capacity, meaning they heat up and cool down more rapidly than water, a substance known for its high specific heat capacity.

This explains why in our exercise, even though both the aluminum pot and the water are heated by the same temperature change, water needs more heat to reach the target temperature. By quantifying this heat using specific heat capacity, we can calculate precise energy demands for a vast array of applications, from culinary endeavors to industrial processes.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In this context, we're particularly interested in the first law of thermodynamics, which posits that energy cannot be created or destroyed, only transformed.

When applying this to our exercise, the energy in the form of heat that we supply to the pot and water is not lost; rather, it is transformed into an increased internal energy of the substances, which we observe as a rise in temperature. The more heat we apply, the more we increase the internal energy, and the higher the temperature goes until we reach our desired temperature of 95.0°C.

By involving thermodynamics in heat calculations, we're not just looking at temperature change; we're delving into the fundamental energy exchanges and transformations that are taking place at the molecular level within the pot and water.
Heat Calculation
Heat calculation is essentially the application of specific heat capacities in thermodynamic equations to ascertain the amount of energy transfer required to achieve a certain temperature change. In our problem, we put heat calculation into practice by using the formula:
\[Q = mc\Delta T\]
where:
  • \(Q\) is the heat energy transferred,
  • \(m\) is the mass of the substance,
  • \(c\) is the specific heat capacity, and
  • \(\Delta T\) is the change in temperature.
Given the initial and final temperatures alongside the masses and specific heat capacities of our substances, we calculated the heat required for both the aluminum pot and the water separately, and then summed these quantities for the total heat required. Clear, step-by-step calculations guide us through complex problems and ensure that students precisely understand energy transfer in thermodynamic processes.

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Most popular questions from this chapter

A 2.0 -kg metal object with a temperature of \(90^{\circ} \mathrm{C}\) is submerged in \(1.0 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C}\). The water-metal system reaches equilibrium at \(32^{\circ} \mathrm{C}\). What is the specific heat of the metal? a) \(0.840 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) b) \(0.129 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) c) \(0.512 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) b) \(0.129 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) d) \(0.433 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\)

A gas has an initial volume of \(2.00 \mathrm{~m}^{3}\). It is expanded to three times its original volume through a process for which \(P=\alpha V^{3},\) with \(\alpha=4.00 \mathrm{~N} / \mathrm{m}^{11} .\) How much work is done by the expanding gas?

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature \(T=300 . \mathrm{K}\), which is approximately the surface temperature of Earth.

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